= Solution
{c}
Let $P$ and $Q$ be <subgroups> of orders five and seven, which exist by <6/c/i>. Their <normalizers> contain $P$ and $Q$, respectively. By <Lagrange's theorem>, $N_G(P)$ has order $5$ or $35$, and $N_G(Q)$ has order $7$ or $35$. If either has order $35$, its <subgroup> is normal by <6/b/iv>.
Suppose neither is normal. Then <6/b/iv> gives at least seven distinct <subgroups> of order five and at least five distinct <subgroups> of order seven. Distinct <subgroups> of the same prime order intersect only in the identity, and a <subgroup> of order five and one of order seven also intersect only in the identity by <Lagrange's theorem>. Thus $G$ would have at least
$$
7(5-1)+5(7-1)=58
$$
nonidentity elements, contradicting $|G|-1=34$. Hence \b[at least one of $P,Q$ is a normal subgroup].
Back to article page
